Plzzzzzz plzzzzz help will give brainlist

If you found a galaxy with an Ha emission line that had a wavelength of 756.3 nm, what would be the galaxy’s distance if the Hub

Delicious77 [7]

Answer: 653.011Mpc

Explanation:

Hubble deduced that the farther the galaxy is, the more redshifted it is in its spectrum, and noted that all galaxies are "moving away from each other with a speed that increases with distance", and enunciated the now called Hubble–Lemaître Law.

This is mathematically expressed as:

$V=H_{o}D$ (1)

Where:

$V$ is the approximate recession velocity of the galaxy

$H_{o}=70 km/s/Mpc$ is the current Hubble constant

$D$ is the galaxy's distance

On the other hand, the equation for the Doppler shift is:

$\frac{\Delta \lambda}{\lambda_{o}}=\frac{V}{c}$ (2)

Where:

$\lambda_{o}=656.3nm=656.3(10)^{-9}m$ is the wavelength for the Ha line of the galaxy observed at rest

$\Delta \lambda=\lambda_{1}-\lambda_{o}$ is the variation between the measured wavelength for the Ha emission line in the spectrum of this galaxy ( $\lambda_{1}=756.3nm=756.3(10)^{-9}m$ in this case) and the wavelength for the same Ha line observed at rest

$c=3(10)^{8}m/s$ is the speed of light

Rewriting (2):

$\frac{\lambda_{1}-\lambda_{o}}{\lambda_{o}}=\frac{V}{c}$ (3)

Isolating $V$ :

$V=\frac{(\lambda_{1}-\lambda_{o})c}{\lambda_{o}}$ (4)

Finding $V$ :

$V=\frac{(756.3(10)^{-9}m-656.3(10)^{-9}m)3(10)^{8}m/s}{656.3(10)^{-9}m}$ (5)

$V=45710802.99m/s=45710.80299km/s$ (6)

Substituting $V$ in (1):

$45710.80299km/s=(70km/s/Mpc)D$ (7)

Finding $D$ :

$D=\frac{V}{H_{o}}=\frac{45710.80299km/s}{70km/s/Mpc}$ (8)

Finally:

$D=653.011Mpc$ (8) This is the galaxy's distance

6 0

3 years ago

DEFINE accelerated motion

posledela

Answer:

accelerated

Explanation:

I think not sure

3 0

2 years ago

Consider a transformer used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 4 turns in it

Ilia_Sergeevich [38]

Answer:

Output Voltage = 0.96 volts

Explanation:

As we know by the equation of transformer we have

$\frac{V_1}{V_2} = \frac{N_1}{N_2}$

here we know that

$V_1$ = voltage of primary coil = 120 Volts

$V_2$ = voltage of secondary coil

$N_1$ = number of turns in primary coil = 500

$N_2$ = number of turns in secondary coil = 4

now we will have

$\frac{120}{V} = \frac{500}{4}$

$V = 0.96 Volts$

4 0

3 years ago

A force of 8 N accelerates by 4 m/s^2. What would be the amount of force needed to give a final acceleration of 5.3 m/s^2

Elis [28]

Answer:

10.6 N

Explanation:

F = ma

8 = m * 4

m = 2 kg

F = 2 * 5.3

F = 10.6 N

4 0

2 years ago

WILL GIVE BRAINLY!! PLEASE HELPP

LekaFEV [45]

The purpose of the machine is to leverage its mechanical advantage such that the force it outputs to move the heavy object is greater than the force required for you to input.

But there's no such thing as a free lunch! When you apply the conservation of energy, the work the machine does on the object will always be equal to (in an ideal machine) or less than the work you input to the machine.

This means that you will apply a lesser force for a longer distance so that the machine can supply a greater force on the object to push it a smaller distance. That is the trade-off of using the machine: it enables you to use a smaller force but at the cost of having to apply that smaller force for a greater distance.

The answer is: The work input required will equal the work output.

8 0

3 years ago